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zavuch27 [327]
2 years ago
10

1. In the image below, the purple particles are protons and the white particles are neutrons. Which of the following equations m

atches the balanced reaction shown in
the diagram?

Physics
1 answer:
lara [203]2 years ago
5 0

The balanced reaction that could lead to the image shown is; 2/1 H + 1/1H -----> 3/2He.

<h3>What is a nuclear equation?</h3>

A  nuclear equation is one which involves changes in the nucleus of atoms to yeild new isotopes of elements.

We know that a nuclear equation must be balanced. The balanced reaction that could lead to the image shown is; 2/1 H + 1/1H -----> 3/2He.

Learn more about nuclear equation:brainly.com/question/19752321

#SPJ1

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Which of the following creates an adhesive force that prevents separation of the parietal and visceral pleurae during ventilatio
Diano4ka-milaya [45]

Answer:

Negative intrapleural pressure is the correct answer

Explanation:

Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.

Air moves from a region of higher pressure to one of lower pressure.  Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.

Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.

3 0
3 years ago
Compared to a stationary galaxy, light from a galaxy that is moving away from earth will appear _____.
Vlad1618 [11]
"B" When an object moves away from us, the light is shifted to the red end of the spectrum, as its wavelengths get longer.
3 0
3 years ago
Read 2 more answers
A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
emmasim [6.3K]

<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

<em>           = 460.9 N</em>

<em>Minimum push required to move the box up the ramp is 460.9 N</em>


7 0
3 years ago
Suppose we repeat the experiment from the video, but this time we use a rocket three times as massive as the one in the video, a
shusha [124]

Answer:

2/3

Explanation:

In the case shown above, the result 2/3 is directly related to the fact that the speed of the rocket is proportional to the ratio between the mass of the fluid and the mass of the rocket.

In the case shown in the question above, the momentum will happen due to the influence of the fluid that is in the rocket, which is proportional to the mass and speed of the same rocket. If we consider the constant speed, this will result in an increase in the momentum of the fluid. Based on this and considering that rocket and fluid has momentum in opposite directions we can make the following calculation:

Rocket speed = rocket momentum / rocket mass.

As we saw in the question above, the mass of the rocket is three times greater than that of the rocket in the video. For this reason, we can conclude that the calculation should be done with the rocket in its initial state and another calculation with its final state:

Initial state: Speed ​​= rocket momentum / rocket mass.

Final state: Speed ​​= 2 rocket momentum / 3 rocket mass. -------------> 2/3

8 0
3 years ago
If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?
aivan3 [116]

Initial velocity = \(v_0\)

acceleration in the downward direction = -9.8 \(\frac {m}{s^2}\)

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

\(v^2 = {v_0}^{2} + 2aH

\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410

\({v_0}^{2} = 2 \times 9.8 \times 0.410\)

\(v_0 = 2.834 \frac {m}{s}\)

Speed with which the flea jumps = \(2.834 \frac {m}{s}\)

4 0
3 years ago
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